$\begingroup$@Wanderer When I first saw the series, I immediately thought of $\sum\limits_{n=1}^\infty\frac1n$, so I tried to find whether $\frac1n\lt\frac1{(\log n)^3}$ holds. But with $n^2\lt(\log n)^3$ we will most probably get nothing since $\sum\limits_{n=1}^\infty\frac1{n^2}$ converges.$\endgroup$
– WorkaholicJan 22 '15 at 20:57